area under and over the curve

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jackie
jackie am 12 Aug. 2018
Kommentiert: Angelavtc am 28 Apr. 2020
hello I hope you can help me!
as per diagram below, I have a curve (blue) and a straight line (pink) say y=1.
I will need to find the unit area under the curve (yellow area) and "over" the curve (green area).
also the curve dont have many data points, I hope this wouldn't be a huge problem!
Please advise and many thanks in advance!

Akzeptierte Antwort

Image Analyst
Image Analyst am 12 Aug. 2018
Try this:
xLast = find(y > 1, 1, 'last');
areaUnderCurve = sum(abs(y(1:xLast) - 1))
This is basically assuming the digitized y is a bar chart and you're summing the heights of the rectangular bars. Of use trapz() if you want trapezoids instead of rectangular bars.
  2 Kommentare
jackie
jackie am 12 Aug. 2018
Bearbeitet: jackie am 12 Aug. 2018
hey Image Analyst, thank you for the prompt reply!
if possible , Can you quickly explain the 2 line of code please if possible?
and also how to use trapz()?
thanks!
Image Analyst
Image Analyst am 12 Aug. 2018
y starts below 1, then rises above 1, then finally drops below 1 again. It's this x coordinate, where it drops below 1 that we need to find because that defines the rightmost place where you want to integrate.
If you do y-1, that will give you the area when y > 1, but that won't work on the left because it will give you negative areas. You'd want 1-y there. But 1-y is simply abs(y-1) so that now allows us to use the same formula everywhere because that will also work for the right area because abs(a positive number) is equal to the positive number.
trapz() is explained in the documentation. Do you really want me to just repeat what's already in the documentation? Basically instead of assuming the areas of each point are rectangular bars, it draws a line from one data point to the adjacent one and this gives you bars with sloping tops rather than flat tops. It's just a different definition and way of doing it. Use whichever you want.

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Weitere Antworten (3)

jackie
jackie am 13 Aug. 2018
Bearbeitet: jackie am 13 Aug. 2018
ok I figured it out, thanks for you help!
x=[0,5,10,20,50,100,150,250,350,450];
b=[4.01E-01,8.30E-01,9.49E-01,1.08E+00,1.21E+00,1.23E+00
,1.20E+00,1.12E+00,1.11E+00,9.69E-01];
line=[1,1,1,1,1,1,1,1,1,1];
x_fine=0:10:450
b_fine = interp1(x,b,x_fine,'spline');
figure
plot(x,b,'o',x_fine,b_fine,'r', x,line,'b');
xFirst = find(b_fine > 1, 1, 'first');
xLast = find(b_fine > 1, 1, 'last');
area_total = trapz(x_fine,b_fine-1)
area_Loss = trapz(x_fine(1:xFirst),b_fine(1:xFirst)-1)
area_Gain = trapz(x_fine(xFirst:xLast),b_fine(xFirst:xLast)-1)
  1 Kommentar
Angelavtc
Angelavtc am 27 Apr. 2020
But using this formula, you are not caculating the yellow area but also all the area that is below y=1

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Image Analyst
Image Analyst am 27 Apr. 2020
This is how I'd do it:
%--------------------------------------------------------------------------------------------------------------------------------------------------------
% CLEAN UP - INITIALIZATION STEPS
clc; % Clear the command window.
fprintf('Beginning to run %s.m.\n', mfilename);
close all; % Close all figures (except those of imtool.)
clearvars; % Erase all existing variables.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;
%--------------------------------------------------------------------------------------------------------------------------------------------------------
% TRAINING DATA PREPARATION
x = [0,5,10,20,50,100,150,250,350,450];
b = [4.01E-01,8.30E-01,9.49E-01,1.08E+00,1.21E+00,1.23E+00,...
1.20E+00,1.12E+00,1.11E+00,9.69E-01];
line = ones(1, length(x));
x_fine = linspace(min(x), max(x), 1000); % a thouisand points.
b_fine = interp1(x,b,x_fine,'spline');
line_fine = ones(1, length(x_fine));
hFig = figure
plot(x, b, 'bo', 'LineWidth', 2);
hold on;
plot(x, line, 'b', 'LineWidth', 2);
plot(x_fine, b_fine, 'r-', 'LineWidth', 2);
grid on;
xlabel('X', 'FontSize', fontSize);
ylabel('b', 'FontSize', fontSize);
title('trapz() demo', 'FontSize', fontSize);
indexLeft = find(b_fine > 1, 1, 'first');
indexRight = find(b_fine > 1, 1, 'last');
% Put up vertical lines there.
xline(x_fine(indexLeft), 'Color', 'm', 'LineWidth', 2);
xline(x_fine(indexRight), 'Color', 'm', 'LineWidth', 2);
%--------------------------------------------------------------------------------------------------------------------------------------------------------
% COMPUTE AREAS BETWEEN CURVE AND FLAT BLUE LINE
% Find that area below the line at 1.
y = line_fine(1 : indexLeft - 1) - b_fine(1 : indexLeft - 1);
area_Loss = trapz(x_fine(1 : indexLeft - 1), y)
% To double check, compute area by summing also.
area_Loss2 = sum(y) * abs(x_fine(2) - x_fine(1)) % Should be fairly close.
str = sprintf('Area below blue line = %.3f', area_Loss);
text(20, 0.86, str, 'Color', [0, 0.5, 0], 'FontSize', fontSize);
% Find that area above the line at 1.
y = b_fine(indexLeft : indexRight) - line_fine(indexLeft : indexRight);
area_Gain = trapz(x_fine(indexLeft : indexRight), b_fine(indexLeft : indexRight)-1)
% To double check, compute area by summing also.
area_Gain2 = sum(y) * abs(x_fine(2) - x_fine(1)) % Should be fairly close.
str = sprintf('Area above blue line = %.3f', area_Gain);
text(140, 1.05, str, 'Color', [0, 0.5, 0], 'FontSize', fontSize);
area_total = area_Loss + area_Gain
str = sprintf('Sum of both areas = %.3f', area_total);
text(220, 0.75, str, 'Color', [0, 0.5, 0], 'FontSize', fontSize);
hFig.WindowState = 'maximized';
fprintf('Done running %s.m.\n', mfilename);

Angelavtc
Angelavtc am 28 Apr. 2020
This is wonderful @Image Analyst, with this, you helped me to answer a question that I posted here: https://www.mathworks.com/matlabcentral/answers/521056-how-to-find-the-are-between-two-graphs?s_tid=mlc_ans_email_view#comment_836228
if you want you can post this answer and I will accept it.
Thank you!
  1 Kommentar
Angelavtc
Angelavtc am 28 Apr. 2020
So basically, the point was to calculate the difference between both areas (blue line vs red spline curve)

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